Dynamics
a branch of mechanics that deals with the “motion”
of bodies under the action of forces.
Two parts:
1. Kinematics
Study of motion without reference to the
forces that cause the motion
2. Kinetics
Study of motion under the action of forces on bodies
for resulting motions.
1
Kinematics
Kinetics
study of objects’ motion
without reference to the
forces that cause the
motion
study of motion under
the action of forces on
bodies
about
the
bodies’ motions.
kinematics
relation is
necessary to solve
kinetics problem
How does wAB related with another wＣＤ
(kinematics)
If you want aCD = 36.5 rad/s2,
how much torques do you apply to AB
(kinetics)
2
Kinematics
Particles
Kinetics
Before Midterm
Rigid Bodies
After Midterm

3
How to describe the motion?
Kinematics of particles
How to describe the position?
 use : Frame
 Frame: Ref-Point + Coordinate
Motion: time-related position
 use: rA (t ) at any tme t
Coordinate
Path of the particle
(x,y) coord
A'
rA'
(r,q) coord
r
rA
q

r
s
A
O Reference
Point
relative frame

= “position vector” (start from
some convenient point: reference point)
time-related info: velocity , acceleration
 from A to A' takes t second
 Distant traveled:
s
 Displacement ( in t ):
n-t coord
v
s
t
measured along the path, scalar
You are going to learn:
(x,y) coord

r
r-q coord

r
r
v
t
4
2/2 Rectilinear Motion (1D-motion)
“displacement, velocity,
acceleration” can be
considered as scalar quantity.
Particle moving on
straight line path.
reference
point
P’: t+t
P: t
s
+s
“average” velocity:
vaverage
reference axis
s
t
“instantaneous” velocity :
v
s
ds

 s
t 0 t
dt
lim
By defining the axis according to
the moving direction
The particle is at point P at
time t and point P at time
t+t with a moving distance
of s.
Positive v is defined in the
same direction as positive s.
i.e.) positive v implies that s is
increasing, and negative v
5
implies that s is deceasing.
Rectilinear Motion
( similarly as
“instantaneous”
acceleration
a
dv
v
dt
or
v
ds
s
dt
)
d 2s
a 2 s
dt
• Eliminating dt, we have
ds
ds

v
dv
dt
( )
a
vdv  ads
or
sds  sds
3 Equations, but only 2 are independent
Positive a is defined in the same direction as positive v (or s).
Ex. positive a implies that the particle is speeding up (accelerating)
negative a implies that the particle is slowing down (decelerating).6
Formula Interpretations of a,v,s,t
a
time :
0
displacement : so
velocity:
time :
t
displacement : s
velocity:
v
vo
where a is the function of
a  ao or a(t ) or a(v) or a( s)
or a(v, s) or a(v, s, t )
(1) Constant acceleration (a= constant). Find v(t), s(t)
v
dv
a
dt

dv   a dt
v0
0
v
ds
v
dt
t

v dt   ds
v0
v
vdv  ads
s

v0
s0
v  v0  at
1 2
s  s0  v0t  at
2
t
v dv 

0
a ds
v2  v02  2a(s  s08)
g 9.81 m

6
6 s2
(constant)
a
vo  2 m/s
v2  v02  2as
5m
1 
v  v  2 g h
6 
v?
dv
a
dt
2
ds
v
dt
vdv  ads
2
0
1

v 2  22  2 9.815
6

v  4.51m / s
Ans
10
t1  t2 
1
(sec)
2
a=g (const)
1
s  s0  v0 (t )  a (t ) 2
2
1 2
s  gt
2
s2
t1
t2
s1 =3
2nd
Time required for 3-m falling.
3  1 2 9.81t12  t1  0.782s
Thus, the 2nd ball has time to fall:
1st
h  s1  s2
 3  0.39  2.61 m
t2  0.782  0.5  0.282 s
Therefore, the 2nd ball travels:
1
2
s2  9.81  0.282   0.39 m
2
11
v0  3 (t  0)
Formula Interpretations of a,v,s,t
a  3t 2  2t
(2) Acceleration given as a function of time,
a = f(t) . Find v(t), s(t)
dv
a
dt
v

v0
ds
v
dt
t
dv   f (t ) dt
vdv  ads
t
v  v0   f (t )dt
0
0
Or by solving the
differential equation:
d 2s
(a  v  s )
 f (t )
2
dt
v(t)
ds  vdt
s(t)
s(t)
v(t)
12
v0  3 (t  0)
Formula Interpretations of a,v,s,t
(3) Acceleration given as a function of velocity,
a = f(v)
Find s(v), t(v)
t
dv
a
dt
v
 dt  
0
v0
s
vdv  ads

s0
v
ds
dt
v
ds 

v0
v

v0
dv
f (v )
a  4  0.03v 2
v(t)
t(v)
v
dv
f (v )
s(v)
s
v dt 

ds
s(t)
s0
13
v0  3 (t  0)
Formula Interpretations of a,v,s,t
a  4   0.1 ln s 
(4) Acceleration as function of displacement:
a = f (s) Find v(s), t(s) Find s(t), v(t), a(t)
vdv  ads
v

s
vdv   f ( s ) ds
vo
t
ds
v
dt
v(s)
so
s

dt  
to
so
1
ds
v( s)
t(s)
s(t)
v(t)
dv
a ,
dt
a(t)
14
Graphical Interpretations of
• The familiar slopes and areas
s-t curve
a, v, s, t
Area under a-t curve from t1-t2
= v(t2) − v(t1)
t2
adt  dv
v2
 adt   dv  v
2
t1
 v1
v1
a-t curve
v-t curve
v2
vdt  ds
s2
 vdt   ds  s
2
v1
 s1
s1
Area under v-t curve from t1-t2
= s(t2) − s(t1)
15
Graphical Interpretations of
a, v, s, t
• Less familiar interpretations
Find a !
v-s curve
a-s curve
Find v !
q
q
t2
 ads
t1
v2
v22 v12
  vdv 

2
2
v1
ads  vdv
a  v tan q
dv
av
ds
16
Given the acceleration-displacement plot as shown.
Determine the velocity when x = 1.4 m, assuming that the velocity is
0.8 m/s at x = 0
ads  vdv
a-s curve
v

 a dx
v dv 
vo
Area under curve
= 0.16 + 0.12 + 0.08 + 0
= 0.36
1.4
v   1.17 m/s
vo : 
v
+ or -?
a is always + in that range
 vf : 
v  1.17 m / s
ANS
 v2 
   0.36
 2  vo
v  v  0.72
2
2
o
v2  0.72  0.64  1.3618
20
vdv  ads
v0
v
a=a(v)
s
Fig1

Case (b)
v
Case (a)
v
 vdv   a0 ds
v0
0


1 2
v  v02  a0 s
2
2

1 
1000 
2
250


0
 
  2 s

2 
3600 

s
d (v 2 )
v 2  a0  kv 2   0 ds
0
vdv  a0 ds
s

vdv  a0  kv2 ds
s  1206 m
s
250
1
 ln( ao  kv 2 ) 
0
2k
s  1268 m
21
Find v(t), s(t) of the mass
if s start at zero and v  v0
a  k 2 s
v
ds
dt
dv
a
dt
vdv  ads

dx
a2  x2
x
 C1
a
 sin 1
(t  0, s  0, v  v0  0)
vdv  ads  (k 2 s)ds
initial condition
(t  0, v  v0 )
v2
k 2s2

 C1
2
2
v   v0 2  k 2 s 2
v0 2
C1 
2
+ or - ?
1 1 ks
sin
 t  C2
k
v0
s
smax 
v0
sin(kt )
k
initial condition
(t  0, s  0)
v
vo

(when t 
)
k
2k
ds
 v0 cos(kt )
dt
vmax  vo (when t  0,

k
,...)
v  v0 2  k 2 s 2
Altenative Solution: Differential equation
s
ds
 v0 2  k 2 s 2
dt
1
v02  k 2 s 2
ds  dt
ds 2
2

(

a

)

k
s
2
dt
ds 2
 k 2s 
0
2
22
dt
v
Find h and v when the
ball hits the ground.
dv
dt
vdv  ads
a
downward:
vo 
vdv  ads  ( g  kv2 )ds
k  0.006
g  9.81
Upward:
vdv  ads  ( g  kv2 )ds
v
 (
)dv  s  h
2
g  kv
0
1
1
g  kv02
2
h   ln( g  kv0 ) 
ln(
)
vo
2k
2k
g
ds
dt
vf
0
v
(
0 kv 2  g )dv  h ds  h
kv f 2  g
kv f 2
vf
1
1
1
2
ln(kv  g ) 
ln(
)
ln(1 
)
0
2k
2k
g
2k
g
vf 
g
(1  e2 kh )
k
= 24.1 m/s
23
= 36.5 m
t  1: s  16
v  18
SP2/1 Given: x(t )  2t 3  24t  6
Find 1) t when v=+72
t  2 : x  26
v0
2) a when v=+32
tv72  4
a(t )  vx  12t
tv32  3
x
How many turns?
At the turn
3) total distance traveled from t=1 to t=4
vx (t )  x  6t 2  24
t  4 : x  38
v  102
tv72  4
v0
t  2 to make v =0
t=-2 is not possible,
*only* 1 U-turn
at 3  36
Correct?
xt 4  xt 1
 38  (16)  54 m
total distance traveled
| xt 2 - xt 1 |  | xt 4 - xt 2 |
net displacement
!= total distance traveled
 | 26  ( 16) |
+ | (38)  ( 26) |
| 10 | + | 64 | 74 m
24
2/44 The electronic trottle control of a
model train is programmed so that the train
speed varies with position as shown.
Determine the time t required for the train
to complete one lap.
Rectilinear motion?
a along the path
not total a
ds
v
dt
dv
a
dt
vdv  ads
v, a
v, a
s
Rectilinear equations can be used for curved motion
if s, v, a are measured along the curve
(more on this soon)
s
25
ds
v
dt
dv
a ,
dt
vdv  ads,
2/44 The electronic trottle control of a
model train is programmed so that the train
speed varies with position as shown.
Determine the time t required for the train
to complete one lap.
C
Area = vds
dv a

ds v
dv

dt
V  V0 (const )
t  2(
T02
2

0.25

Slope = dv/ds
( 0)
2
= Slope = Constant. =C
a  Cv
1
1 0.125
0.125
t  ln(v )0.25  ln
C
C
0.25
2


4

ln 2 
ln 2) 
(1   ln 2)
0.25 0.25
0.25
0.25
= 50.8 s
0.125  0.25
dt 
t 
1 1
( dv )
C v
1


 0.125  0.25 





2


T
ln

2(2 )
2
0.125


ln 2
0.25 0.25

 ln 2
260.25
Recommended Problem
a  g  cy
2/29 2/36 2/46 2/58
v  c0  c1t  c2t 2  c3t 3
slope =0 at
both end
a  kv 2
27
Kinematics of particles
How to describe this
particle’s motion?
 Framework for describing the motion
Reference
Frame
Path of the particle
(x,y) coord
A'
rA'
(r,q) coord
r
rA
q

r
s
A
 Recording rA (t )

r

= “position vector” (start from
some convenient point: reference point)
"time"-related info
 from A to A' takes t second
 displacement ( in t ):

r
O Reference
Point
 Distant traveled:
You are going
to learn:
relative frame
s
v
r
t
v
s
t
measured along the path, scalar
29
(x,y) coord
n-t coord
r-q coord
2/3 Plane Curvilinear Motion
 Motion of a particle along a curved path in a single plane (2D curve)
Reference Frame
Path of the particle
(x,y) coord

r

= “position vector” (start from
some convenient point: reference point)
 from A to A' takes t second


r  r
(r,q) coord
r

r
q
A'

r
s
A
 displacement of the particle
during time t :  r
 Distant traveled = s, measured along
the path, scalar
O
 Basic Concept: “time derivative of a vector”



r

r
v  lim
vavg . 
t 0 t
t
“Average Velocity”

 dr 
v
r
dt
“(Instantaneous) Velocity”
30
Speed and Velocity
“(Instantaneous) Velocity”

 dr 
v
r
dt
Path of the particle
A'


r  r 
 r s
r
A
“(Instantaneous) Speed”
vs
O
t  0 :
s  |r |
ds
dt
s
 1


lim
|

r
|


t 0 t
t 0 t


 lim
Path of the particle

r
v
 lim
t 0
r
r
dr
 lim

v
t 0 t
t
dt
A
O
Velocity vector is tangent to the curve path
Instantaneous Speed
Path of the particle
A'


r  r 
 r s

r
A
O

 dr 
v
r
dt
vs
ds
dt
ds
| v |
dt
dr ds

dt dt
dr d | r |

dt
dt
speed
dr
r
dt
Time rate of change of
length of the position

vector r
(r,q) coord
r
q
32
Acceleration

v2
Path of the
particle
A'
r2
O
Hodograph

v1
A
r1

aavg .

v

t
“Average acceleration”

v2

v

v1
Hodograph

a2

v2

a1

v1
C
C


v
a  lim
t 0 t

 dv 
a
v
dt
“(Instantaneous) Acceleration”
*** The velocity is always tangent to the path of the particle (frequently used in
problems) while the acceleration is tangent to the hodograph (not very
33
important) ***
Vector Equation and
reference frame

 dr 
v
r
dt
Path of the particle
Reference
Frame

 dv 
a
v
dt
A'
Vector equation is in general form, not
depending on used coordinate.
Reference
Frame
(coordinate)
- Rectangular: x-y
rectangular
O
Usage will depend on selection.
More than one can be used
At the same time.
- Normal-Tangent: n-t
- Polar: r-q
r  xiˆ  yjˆ


r  r 
 r s

r
A
r  reˆr
v  veˆt
r-q
n-t
34
Derivatives of Vectors
Derivatives of Vectors: Obey the same rules as they do for scalars
d (uP )
 uP  uP
dt
d ( P  Q)
 P Q  P Q
dt
d ( P  Q)
 P Q  P Q
dt
35
2/4 Rectangular Coordinates
Reference Frame
O
Path

v
ˆj
y
y  vy
Both “divided” particles,
are moving in rectilinear
motion
A( x, y )
y

r
O
y  ay
iˆ
O
x
x
r  x iˆ  y ˆj

 dv 
a
v
dt
x  vx x  ax
Correct?
0
Basic Agreement:
0
v  r  vxx iˆ vyy ˆj  ( xiˆ  yjˆ)
a  v  axx iˆ  ayy ˆj

 dr 
v
r
dt
Direction of
reference axis {x,y} do
not change on time
variation.
Rectilinear Motion in 2 perpendicular & independent axes.
Path
ˆj
y
ry  y
O
vy  y

v
A( x, y )
vx  x

r
rx  x

 dr 
v
r
dt
iˆ
ds
dt
dv
a
dt
vdv  ads
ax  x

r
iˆ
x
O
x
ay  y
y

 dv 
a
v
dt
t
v

a
ˆj
rectilinear in 2 dimension,
related which other via time.
y
x
vx
vy
ax
ay
rectilinear
in x-axis
rectilinear
in y-axis
If given
ax  f ( x, vx , t )
ay  f ( y, vy , t )
can you find
vx  vx (t )
v y  v y (t )
sx  sx (t )
s y  s y (t )
Common Cases

r  xiˆ  yˆj
 
v  r  xiˆ  yˆj
 
a  v  xiˆ  yˆj
Rectangular coordinates are usually good
for problems where x and y variables
can be calculated independently!
Ex1) Given ax = f1(t) and ay = f2(t)
ax  f1 (t )
a y  f 2 (t )
t
v x   f1 (t )dt
o
t
v y   f 2 (t )dt
o
a  f 2 (t )
From this, you can find
“path” of particles
t
s x   v x (t )dt
o
t
s y   v y (t )dt
o
Ex2) Given x = f1(t) and y = f2(t)
sx  f1 (t )
a  f1 (t )
s y  f 2 (t )
vx  f1(t )
v y  f 2(t )
a x  f1 (t )
a y  f 2 (t )
Projectile motion
•
The most common case is when ax = 0 and ay = -g
(approximation)
• x and y direction can be calculated independently
v
ds
dt
a
dv
ads  vdv
dt
Note: a = const
y
vo
vx
vx
(vo)y =
v0 sinq
v  vo  at
a=0
v
vy
g
q
vy
v
(vo)x = v0 cosq
x-axis
vx = (vo)x
x = xo + (vo)xt
x
y-axis
v y  (vo ) y  gt
vy 2
1
s  so  vot  at 2
2
(v ) 

x - xo
eliminate t  t =
, tan q  o y 
(vo )x
(vo ) x 

y  yo  (tan q )( x  xo ) 
1 2
gt
2
 (vo )2y  2g( y  yo )
y  yo  ( vo ) y t 
v2  vo2  2a(s  so )
g
2
(
x

x
)
o
2v02 cos2 q
y  (tan q ) x 
( in case of
g
2
x
2v02 cos2 q40
xo  yo  0 )
y
Determine the minimum horizontal velocity u that a boy
can throw a rock at A to just pass B.
x
Note: rectilinear (a=const)
v  vo  at
g
(1)
(2)
v2  vo2  2a(s  so )
1
(3)
s  s o  v o t  at 2
2
it can be applied in both x and y
direction
1
x  xo  vx t  a x t 2
2
ax  0
x  ut
vx  u xo  0
40  u (1.43)
u  28
m/sec.
If we use eq. (3) in the y-direction :
1
y  yo  v y t  a y t 2
2
1
1
y   gt 2
 10   (9.81) t 2
2
2
t  1.43
sec.
aD  kv
axiˆ  a y ˆj
k : const
k (vxiˆ  v y ˆj )
Find x-,y-component of velocity and
displacement as function of time , if
the drag on the projectile results in an
acceleration term as specified. Include
the gravitational acceleration.
ˆ  (kv y  g ) ˆj


kv
i
ˆ
x
a  aD  gj
vy
dv x
 a x   kv x
dt
vx  vx0 e
t
x
0
 kt
vx

v x0
dv y
1
dv x  kt
vx
 v0 cosq e
dt
v cos q
v0 cos q e dt  0
(1  e  kt )
k
vx  0
xmax 
v y0
g
vy 
k
dv y  kt
g  kt g
g  kt g
v y  (v y0  )e   (v 0 sin q  )e 
k
k
k
k
 kt
g
g

y   (v 0 sin q  )e  kt   dt
k
k
0 
1
g
g
 (v 0 sin q  )(1  e  kt )  t
k
k
k
t
 kt
t  :
 a y  kv y  g

1
vy  
v 0 cos q
k
g
k
y  
42
Find R=R(q)
2/95 Find q which maximizes R
(in term of v-zero and a)
y
y  ( v0 sin q )t 
x
R sin a  (v0 sin q )(
x  (v0 cosq )t
R=R(q)
R cosa  (v0 cosq )tB
Find Rmax
R cos a
tB 
v0 cos q
1 
q  ( a)
 dR

 dq  0
R cos a
1 R cos a 2
)  g(
)
vo cos q
2 vo cos q
2v02 cos2 q
R
(tan q  tan a )
g cos a
2v02
( 2cos q sin q )(tan q  tan a )  cos2 q (sec 2 q )  0

g cos a
(sin2q )(tana  tanq )  1  0
sin 2q (tan a )  (1  cos 2q )  1  0
2 2
2q  tan 1 ( 
a
1 2
gt
2
a  0  q max 

2
(sin 2q )(tan a )  2sin2 q  1  0
tan 2q  
1
tan a
1
1


)    tan 1 (
)    ( a)   a
tan a
tan a
2
2
43
( v0 : fixed)
H12-96 A boy throws 2 balls into
the air with a speed v0 at the
different angles {q1, q2} (q1  q2).
If he want the two ball collide in the
mid air, what is the time delay
between the 1st throw and 2nd
throw.
The first throw should be q1 or q2?
y  (tan q ) x 
g
x2
2
2
2v0 cos q
g
y  (tan q1 ) x  2
x2
2
2v 0 cos q1
y  (tan q2 ) x 
g
x2
2
2
2v0 cos q2
q1
xc
t1 
v0 cos q1
t2 
t  t1  t2 
x  0,
xc 
intersected point
2v (tan q1  tan q 2 )
g ( 1  1 )
cos2 q1 cos2 q 2
2
o
2vo2 sin(q1  q2 )cosq1 cosq2
xc =
g
(cos2 q2  cos2 q1 )
xc
v0 cos q2
xc
1
1
(

)
v0 cos q1 cos q2

xc cos q2  cos q1
(
)
v0 cos q1 cos q2
=
2vo sin(q1  q2 )
g (cos q1  cos q2 )
44
2/84 Determine the maximum horizontal range R of the projectile
and the corresponding launch angle q.
x  (v0 cosq )t
y  ( v0 sin q )t 
this way?
Yes! but why?
How do you throw the ball?
0   v0 sin q   2(  g )h
2
2
hq 45o  15.928 m
Ceil’s height (5 m) is
our limitation!
R(q)
q=45 ?
h(q)
v sin q
h
2g
2
o
2
2v02
R
(sin q )(cos q )
g
v02
 sin 2q
g
y  (tan q ) x 
1 2
gt
2
g
x2
2
2
2v0 cos q
h
R = R(h)
Range : 0  q 

4
q  h R
qR  qh
max
max
46
2/84 Determine the maximum horizontal range R of the projectile
and the corresponding launch angle q.
x  (v0 cosq )t
y  ( v0 sin q )t 
this way?
v sin q
h
2g
2
o
2
2v02
R
(sin q )(cos q )
g
v02
 sin 2q
g
q  sin
1
2 gh
vo
y  (tan q ) x 
1 2
gt
2
g
x2
2
2
2v0 cos q
q  23.3o
to make h=5
Rq 23.3o  46.3m
47
H12/92 The man throws a ball with a speed v=15 m/s. Determine the
angle q at which he should released the ball so that it strikes the wall
at the highest point possible. The room has a ceiling height of 6m.
48
2/5 Normal and Tangential Coordinates (n-t)
Curves can be considered as
many tangential circular arcs
Fixed point
on curve
t
Path: known
s
O
n
O
t
n
n
O
t
 takes positive t in the direction of
increasing s
 and positive n toward the center of the
curvature of the path
 the origin and the axes move (and
rotate) along with the path of particle
Forward velocity and forward acceleration
make more sense to the driver
The driver is only aware of forward
direction (t) and lateral direction (n).
Brake and acceleration force are often more convenient to describe relative to the
car (t-direction).
Turning (side) force also easier to describe relative to the car (n-direction)
50
Normal and Tangential Coordinates (n-t)
generally,
not total a
Rectilinear Similarity
s
n
ds
dt
v,a
a
dv
dt
s
O
n
n
O
t
eˆn
v
t
Path: known
O
t
eˆt
The reason why we define
this coordinate
0 why?
v  vt eˆt  vneˆn
v  v  vt eˆt  vt
vt , at
s
vdv  ads
Consider: scalar variables (s) along the path (t direction)
similar to rectilinear motion
vt  s  v

v  veˆt
at  vt  s
a  at eˆt ?
s measured along the path
a  veˆt  ( )eˆ51n
Velocity
Small curves can be
considered as circular arcs
** The velocity is always tangent to the path **
Path
C
Velocity ( v ):
Speed:

A
d
eˆn
A
Fixed point
on curve

v
eˆt
ds
v  vt  s 
dt

v  veˆt  ( ) eˆt
ds =  (d)
y  f ( x)
y
path
x

  df 
1   
 dx 



d2 f
dx 2
d
 
dt
(v   )
2





3
2
52


(v   ) v  veˆt  eˆt
Acceleration
Path
eˆt

v
C
 eˆ
n
d
A
n
eˆn
deˆn
d
eˆn
eˆn
t
Need: eˆt 
deˆt
dt
(similarly)
deˆt  (| eˆt | d  )eˆn
deˆn
   eˆt
dt
eˆt   eˆn
ds = (d)
A
n
dv
d ( veˆt )

 veˆt  veˆt
a
dt
dt
 eˆt
v
d
a  veˆt  ( )eˆn
eˆt
d
deˆt
eˆt
t
a  veˆt  v eˆn  veˆt 
v2

eˆn
at  v  s
2
v
2
a

v




0 n

a  at2 53an2
Alternative Proof of eˆt
Using x-y coordinate
Path
eˆn  (sin  )iˆ  (cos ) ˆj
y
C


n
eˆn

v
eˆt
t
eˆt  (cos )iˆ  (sin  ) ˆj
deˆt
 (  sin  )iˆ  (  cos  ) ˆj
dt
 {( sin  )iˆ  (cos  ) ˆj}
A
O
  eˆn
x
deˆt
 eˆn
dt
deˆn
   eˆt
dt
54
Understanding the equation
Path
eˆt
n

v
C
 eˆ
n
n
d
eˆn
A
A
 eˆt
v
a  veˆt  v eˆn
t
ds = (d)

| (dv )n | vd

an | (dv )n | / dt  v
an comes from changes in the

direction of v

an

dv 
at
t

(dv ) n

(dv )t
 d
v 
v dv  (dv )  (dv )
t
n
| (dv )t | eˆt  | (dv )n | eˆn
| (dv )t | dv
| ( dv )t | dv
at 

vs
dt
dt
at comes from changes
 in55
the magnitude of v
What to remember
v  vt eˆt  vneˆn
by definition
of n-t axis

a  at eˆt  an eˆn
2
v
ds
dt

a
dv
dt

 dv
d (veˆt )
a
 v eˆt  v

dt
dt
 v eˆt  v  eˆn
 
   eˆn

 veˆt 
generally,
not total a
Rectilinear Similarity

v
a  v eˆt 
eˆn
 deˆt

 dt
v  vt eˆt  v eˆt
 deˆt 


dt


v
v,a

eˆn
at 
s
s
vdv  ads
y  f ( x)
2
y
path
x
v  
vt 
vt , at
ds
dt
dvt
dt
vt dvt  at ds

  df 
1   
  dx 

d2 f
dx 2
2





3
2
v  veˆt
Proof
(v   )
y


 dy 
1




dx




d2y
dx 2
path
2
x

v

a  veˆt  veˆn  veˆt  eˆn
2





3
2

2
ds 
 (d  )  ds
ds

d
y

d
ds

dx
2
1
2
ds 3
1
dy

 3
 2 ( ) 
(
1

)


2
d y dx
d y
 dx 
dx 2
dx 2
dy
tan  
dx
d
 dy 
dx  dy  ( 1  
 )dx
 dx 
2
dy
d2y
(sec  )d   d ( )  2 dx
dx
dx
2
dy
2
x
d2y
d 2 y  dx 
2
d 
(cos  )dx 
 dx
2 
2
dx  ds 
dx
57
Direction of vt an
v
a  veˆt 
2

an is always plus. Its direction is toward
the center of curvature.
eˆn
B
Speed
B
Speed
Decreasing
Increasing
Inflection point
Inflection point
v
a
A
v
a
A
59
n-t coordinates are usually good for problems
where
“ Curvature path is known ”
vt , at
1) distance along the curvature path (s)
is concerned
2) curvature radius () is
concerned.

v  veˆt
a  veˆt 
v
2

eˆn
s
vt  s  v
at  v  s
v  s  
an 
v2

 v   2
60
A rocket is traveling above the atmosphere such that g = 8.43 m/sec2.
However because of thrust, the rocket has an additional acceleration
component a1 = 8.80 m/sec2 and the velocity v = 8333.33 m/sec. Compute
the radius of curvature  and the rate of change of the speed
 an
Find 
Here :
t
Find
v
 at
a n  g cos60  (8.43) (0.5)  4.215
an 

a1
v2

4.215 
8333.332

  16,475,537 m  16,470 km.
30°
v

60°
an
 a t  a1  g cos30
=> 
 8.80  (8.43) cos30
v
n
g = 8.43 m/sec2
 1.5 m / sec2
ANS
62
at , an , a ?
The driver applies her brakes to produce a uniform deaccceleration.
Her speed is 27.8 m/s at A and 13.89 m/s at C. She experience a
acceleration of 3 m/s^2 at A. Calculate a , a , a ?
t
n
1) the radius of curvature at A
2) the acceleration at the inflection point B a  3
3) the total acceleration at C
at : const
 vdv   a ds  a  ds
t
at 
Condition at A:
t
1 2
( vC  v A2 )  2.41
2s
an 2  a 2  at 2  1.785
v 2 (27.8)2
 
 432m
an
1.785
Condition at B:
Condition at C:
at  2.41 an  0
at  2.41
v2
(13.89)2
an 

 1.286

150
63
Circular Motion (special case)
vt  s  v
direction n? t?
at  v  s
t
v
at
r

v  veˆt
v  s  
a  veˆt 
n
v2
an
q
Particle is moving clockwise
with speed increasing.

eˆn
an 
v2

 v   2
  r (const.) q  
v
 rq
a t  rq
v2
2


r
q
 vq
an r
64
2/123 Determine the velocity and the acceleration
( y, y ) of guide C for a given value of angle q if
C and P shares the
vertical velocity,
acceleration.
(I) q  w
t
n
q
q 0
y  v sin q  rw sin q
v  rq  rw
at  0
an  rw2
q
q
y  an cos q  at sin q
 rw 2 cos q
v  rq
at ( v)  rq
2
v
an (  )  rq 2
r
(II) q  0 q  a
an  0
v  rq  0
at  ra
q q
y  v sin q  0
y  an cos q  at sin q
 ra sin q
65
dy
x
dx
The motorcycle starts from A
with speed 1 m/s, and increased
its speed along the curve at
t

t0
v 1
s0
ds

dt
6.25
x |t 5  ....
t 5
v  ...
v  0.1 m/s2 (const.)
Determine its velocity and
acceleration at the instant t = 5s .
dy
tan  
dx
v  vt  1  (0.1)t
0.1
s  t  ( )t 2
2
d
( vt )  at 
dt
s |t 5  6.25 m
v |t 5  1.5 m/s
  tan1( x |t 5 )
sC  6.25
...  xC  ...
3.184
2
 dy 
ds  (dx )2  (dy ) 2  1    dx
 dx 
s
Numerical
Method
what is x , when t = 5 ?
 |t 5  tan 1 (3.184)
 72.564o
x |t 5  3.184
0
1
2
2
1  x dx  ( x 1  x  ln( x  1  x ))  C1
2
s=0, x=0
2
velocity is the vector
(magnitude + direction)!
66
dy
x
dx
n
t

6.25
The motorcycle starts from A
with speed 1 m/s, and increased
its speed along the curve at
at  v  0.1 m/s2 (const.)

Determine its velocity and
acceleration at the instant t = 5s .
xc  3.184
a  at eˆt  aneˆn
0.1
an 
v2

v |t 5  vt |t 5  1.5 m/s
vt  1  (0.1)t
 |t 5  72.564o
1.52
an 
 0.061
37.171

  dy 
1   
 dx 



d2y
dx 2
2





3
2
3
2 2
 |x  x  (1  3.184 )
c
 37.171
3
2 2
(1  x )

1
67
2/6 Polar Coordinates ( r - q )
Radar Coordinate
t
r
q
0.0
0.1
0.2
0.3
0.4
0.5
25.1
26.2
26.1
24.8
23.2
25.2
32.0
35.0
39.0
40.0
37.0
35.0
r
q
Detect
q
r, q , r, q
r
Find: v , a
What is the velocity and acceleration of the plane?
q
 direction of eˆr = direction of positive r
Path
 direction of eˆq = direction of positive q
eˆq
r
A
eˆr
q
reference point
reference axis
r
r  reˆr
eˆr  (?) eˆr  (?) eˆq
dr
v
 r eˆr  r eˆr
dt
dv
a
dt
eˆq  eˆr   eˆ69q
Polar Coordinates ( r - q )
 direction of eˆr = direction of positive r
Path
q
 direction of eˆq = direction of positive q
eˆq
r
r
eˆr
A
r  reˆr
dr
v
 r eˆr  r eˆr
dt
q
reference axis
reference point
a
dv
dt
eˆq  eˆr   eˆq
q
deˆq
-r
eˆr  (?) eˆr  (?) eˆq
eˆq
eˆr
eˆq
dq
dq
deˆr
eˆr
deˆr  (1 dq )eˆq deˆq  (1 dq )(eˆr )
er  q eˆq
eq   q eˆr
70
q
r
Velocity and Acceleration
eˆr  q eˆq , eˆq   q eˆr

 dr
v
 reˆr  reˆr
dt
v  r eˆr  rq eˆq
vr  r
vq  rq
……….. the change of
 the
length of the vector r
………. therotation of
the vector r
v  vr2  vq2
dv
a
 r eˆr  r eˆr
dt
 rq eˆq  rq eˆq
 rq eˆq

a  (r  rq2 )eˆr  (rq  2rq)eˆq
a  ar2  aq2
ar  r  rq 2
aq  rq  2rq
1d 2
(r q )
r dt
Physical meaning
will be discussed
next page 71
Understanding the acceleration equation

a  (r  rq2 )eˆr  (rq  2rq)eˆq
 Magnitude change of
q
d (rq)
r (in r direction)
rdq
r
vq  rq
rqdq
dq
vq
v r
dr
dq
vr  r
Let’s look at how the velocities change

v  reˆr  rqeˆq
v r  dr
 Direction change of
v r  rdq
rq (in q direction)
 Magnitude change of
vq  d (rq )
rq  rq (in q direction)
Direction change of
vq  rq dq
rq 2 (in -r direction)
72
Radar Coordinate
t
r
q
0.0
0.1
0.2
0.3
0.4
0.5
25.1
26.2
26.1
24.8
23.2
25.2
32.0
35.0
39.0
40.0
37.0
35.0
r
q
Detect
q
r, q , r, q
r
Find: v , a
What is the velocity and acceleration of the plane?
v  r eˆr  rq eˆq

a  (r  rq2 )eˆr  (rq  2rq)eˆq
Detect
v, a
Find: r, q , r, q
Circular Motion (Special Case)

v  reˆr  rqeˆq

a  (r  rq2 )eˆr  (rq  2rq)eˆq
r = const
q
vq
r
aq
r
vr  r  0
v  rq
vr
q
ar
ar  r  rq 2  rq 2
a  rq
q
q
t
n
eˆn  eˆr
eˆt  eˆq
75
r-q coord
n-t coord
• direction depends on its
curvature path.
Usually, Path need to be known

v  veˆt
a  veˆt 
v
no
2

eˆn
vt  s
at  vt  s
r
• r-q coord depend on
{ ref point, ref axis }

r  reˆr

v  reˆr  rqeˆq

a  (r  rq2 )eˆr  (rq  2rq)eˆq
vt , at
s
at // vt  v // path (tangent line)
dvr
ar 
dt
dvq
aq 
dt
76

r  reˆr

v  reˆr  rqeˆq
Problem types

a  (r  rq2 )eˆr  (rq  2rq)eˆq
r  r(t ),q  q (t )
r , v, a
instant t o: r, r, r,q ,q ,q
r , v , a (at t=t o )
instant t o: r, r, r,q ,q ,q
r , v , a (at t=t o )
77
r  r(t ),q  q (t )
eˆr  (cos q )eˆx  (sin q ) eˆ y
eˆq  (  sin q )eˆx  (cos q ) eˆ y
r , v, a
2/145 The angular position of the arm is given by the shown function,
where q is in radians and t is in seconds.
The slider is at r = 1.6 m (t = 0) and is drawn inward at the constant rate
of 0.2 m/s. Determine the magnitude and direction (expressed by the
angle relative to the x-axis) of the velocity and acceleration of the slider
t
when t = 4.
1
t2
q  0.8 
q  0.8t 
20
r  0 .8
q  2 .4
eˆr
eˆq

q
r  0.2
q  0.4
r  0.0
q  0.1
v  (r)eˆr  (rq )eˆq  0.2eˆr  0.32eˆq
v
y
10
r 0
r  0.2
r  1.6  0.2t
At t = 4s:
q 
10
a
q
x
  tan 1
0.32
 302o
0.2



a v  tan 1

0.32
180
Ans
 (2.4)
 260o
0.2

a  r  rq 2 eˆr  rq  2rq eˆq
  tan 1
 0.128eˆr  0.24eˆq
0.24
180
 2420   242o  (2.4)
 310.75
0.12

Ans
v  r eˆr  rq eˆq
instant t o: r, r, r,q ,q ,q
r , v , a (at t=t o )
a  (r  rq 2 ) eˆr  (rq  2rq ) eˆq
At the bottom of loop, airplane P has
a horizontal velocity of 600 km/h and
no horizontal acceleration. The
radius of curvature of loop is 1200 m.
determine the record value of
for this instant.
r ,q
eˆn
a
Use n-t coord
to find v,a
v /
2
eˆt
v
eˆq
q
q  tan 1
600
 166.7 m/s
3.6
r  4002  10002
q
a
v2

 23.1 m/s
vq v sin q

 0.575 rad/s
r
r
r  vr  v cosq  154.7 m/s
v
a sinq ( ar )  r  rq
400
 21.8o
1000
q
eˆr
a
q
v
2
a cosq ( aq )  rq  2rq
r  a sin q  rq  12.15 m/s2
q
81
a cos q  2rq
 0.0365 rad/s2
r
The piston of the hydraulic cylinder gives pin A a constant velocity v = 1.5
m/s in the direction shown for an interval of its motion. For the instant when
q = 60°, determine r, r, q , and q where r  OA
r-q coord
2D vector equation
x-y coord
vA  vA
q v
q
vr  r   v cosq
= 60°
vq  r q
v
q
 v sin q
q
r
vq
vr  r
vr  rq
A
vr

O q  60
150 mm
q  60

q  1.5sin 60  1.299 m/s
r-q coord
From viewpoint of
A
vr
O
r  -1.5cos 60o  - 0.75 m / s
r
From viewpoint of
v
x-y coord
ar  r  rq 2
v  1.5iˆ
aq  2rq  rq
a 0
150 mm
The piston of the hydraulic cylinder gives pin A a constant velocity v = 1.5
m/s in the direction shown for an interval of its motion. For the instant when
q = 60°, determine r, r, q , and q where r  OA
2D vector equation
x-y coord
r-q coord
aA  aA
ar  r  rq 2  0

0.15
(7.5) 2  9.74 m / s 2
sin 60
q
r
vq
q 
2 rq
r

vr
O
q  60
150 mm
vr  r

ar  r  rq
q
A
ar
O
q  60
150 mm

From viewpoint of
v
vr  rq
A
r
2( 0.75)(7.5)
 65.0 rad / s 2
0.15 sin 60
r-q coord
From viewpoint of
a=0
aq  2rq  rq  0
= 60°
r  rq 2
q a
q
2
aq  2rq  rq
x-y coord
v  1.5iˆ
a 0
The piston of the hydraulic cylinder has a constant velocity v = 1.5 m/s
For the instant when q = 60°, determine r, r, q , and q where r  OA
= 60°
= 60°
vA  1.5iˆ
r-q coord
vˆA, x  1.5iˆ a A, x  0
aA  0
Point A : vA  1.5iˆ aA  0 ?
x-y coord
vA  vA
mag?
mag?
reˆr  rq eˆq  1.5iˆ 
ˆj
mag?
mag=0
Only max 2 unknown to be solved
(also, r  0)
r
0
q
r-q coord
x-y coord
velocity
mag?
x-y coord
= 60°
1.5iˆ 
ˆj
= reˆr  rq eˆq
eˆr  (cosq )iˆ  (sinq ) ˆj
  rq cosq  1.732
Acceleration
mag?
x-y coord
mag?
ˆj
mag?
r-q coord
q
mag?
q
r-q coord
Alternate Solution
1.5
q
 10
r sin q
eˆq  ( sinq )iˆ  (cosq ) ˆj
mag?
1.5
 vq  rq
sin q
?
q
?
1.5
= (r  rq 2 )eˆr  (rq  2rq )eˆr
q : 2 unknowns
85
86
sx
y
r
q
x-y coord
At t = 0.5 s
sx  (v0 cos30o )t
s y  (v0 sin 30o )t 
1 2 = 6.274 m
gt
2
vx  v0 cos30
o
vy  v0 sin30o  gt
ax  0
ay  g
= 12.990 m
= 25.981 m/s
= 10.095 m/s
vy
sy v x
r  s   sy  2
2
x
q  tan
2
1
sy  2
sx
x
2
= 15.401 m
= 32.495 deg
r  vr  vx cosq  vy sinq
= 27.337 m/s
rq  vq    v x sin q  v y cos q 
q
= -0.353 rad/s
r  rq 2  ar  ax cosq  a y sin q
r  3.35m / s 2
rq  2rq  aq  ax cosq  a y sin q
q  0.717 rad/s
87
2
The slotted link is pinned at O, and as a result of the constant angular velocity 3 rad/s, it
drives the peg P for a short distance along the spiral guide r = 0.4 q m, where q is in
radians. Determine the velocity and acceleration of the particle at the instant it leaves
the slot in the link, i.e. when r = 0.5 m
Use r-q where reference-origin is at O,
and reference axis is horizontal line.
v  (r)eˆr  (rq )eˆq




a  r  rq 2 eˆr  2rq  rq eˆq
Constrained motion
at r=0.5 v  (0.12)eˆr  (0.5  0.3)eˆq
a  (0  0.5  0.32 )eˆr  (2  0.12  0.3  0.5  0)eˆq
r  0.4q
r  0.4q
r  0.4(0.3)  0.12 (for all time)
r  0.4q
r  0.4(0)  0 (for all time)
89
Summary: Three Coordinates (Tool)
Velocity
Reference
Frame
Acceleration
vy
vn
r
(n,t) coord
velocity meter
q
(r,q)
coord
vx  x
at
ar
an
vx
x
y
r
Observer
Path
ay
aq
O
(x,y)
coord
r
Reference
Frame
vr
vq
x
Observer’s
measuring
tool
vt
Path
Observer
y
ax
vy  y
ax  x
ay  y
vn  0
vt  v
v2
at  v
vr  r
vq  rq
an 

ar  r  rq 2
aq  2rq 92
 rq
Choice of Coordinates
Velocity
Reference
Frame
Acceleration
vy
vn
r
(n,t) coord
velocity meter
q
(r,q)
coord
vx  x
at
ar
an
vx
x
y
r
Observer
Path
ay
aq
O
(x,y)
coord
r
Reference
Frame
vr
vq
x
Observer’s
measuring
tool
vt
Path
Observer
y
ax
vy  y
ax  x
ay  y
vn  0
vt  v
v2
at  v
vr  r
vq  rq
an 

ar  r  rq 2
aq  2rq 93
 rq
Translating
Observer
“Translating-only Frame”
will be studied today
No!
Observer’s
Measuring tool
(x,y)
coord
Path
Observer B
(moving)
(n,t) coord
velocity meter
r
q
(r,q)
coord
Rotating
Two observers (moving and
not moving) see the particle
moving the same way?
Observer O
(non-moving)
Which observer sees
the “true” velocity?
A
both! It’s matter of viewpoint.
This particle
path, depends
on specific
observer’s
viewpoint
“relative” “absolute”
Two observers (rotating and non
rotating) see the particle moving
the same way?
Point: if O
understand B’s
motion, he can
describe the velocity
which B sees.
No!
Observer
(non-rotating)
“Rotating axis”
will be studied later.
“translating”
“rotating” 95
2/8 Relative Motion (Translating axises)
 Sometimes it is convenient to describe motions of a particle “relative” to
a moving “reference frame” (reference observer B)
 If motions of the reference axis is known, then “absolute motion” of the
particle can also be found.
 A = a particle to be studied
Reference frame O
Reference frame B
A

rA

rB

rA / B
B
O
frame work O is considered
as fixed (non-moving)
 B = a “(moving) observer”
 Motions of A measured by the observer
at B is called the “relative motions of A
with respect to B”
 Motions of A measured using framework
O is called the “absolute motions”
 For most engineering problems, O attached
to the earth surface may be assumed “fixed”;
96
i.e. non-moving.
Relative position
Jˆ
ˆj
Y
If the observer at B use the x-y **
coordinate system to describe the
position vector of A we have
y
A

rA

rA / B

rB
O
x
iˆ

rA/ B  xiˆ  yˆj
B
X
where
Here we will consider only the case
where the x-y axis is not rotating
(translate only)
Iˆ

rA / B = position vector of A relative to B (or with respect to B),
iˆ and ˆj are the unit vectors along x and y axes
(x, y) is the coordinate of A measured in x-y frame
** other coordinates systems can be used; e.g. n-t.
97
Relative Motion (Translating Only)
ˆj
y
 x-y frame is not rotating
(translate only)
A
Y

rA
rA / B

rB
x
O
B
X
iˆ  0
ˆj  0
iˆ
rA  rB  rA / B
xiˆ  yjˆ
Note: Any 3 coords
can be applied to
Both 2 frames.

 
a A  aB  a A / B
Direction of frame’s
unit vectors do not
change
0
rA  rB  ( xiˆ  yjˆ )  ( xiˆ  yjˆ)
vA / B
Notation using when
B is a translating frame.
  
vA  vB  vA / B
rA  rB  xiˆ  yjˆ  ( xiˆ  yjˆ98
)0
aA / B
Path
Understanding the equation
Translation-only Frame!
Observer B
A
O & B has a “relative” translation-only motion
  
vA  vB  vA / B
This particle
path, depends
on specific
observer’s
viewpoint
Observer O
reference
reference
framework O
vA / O
frame work B
vB / O
A

rA

rB
O

rA / B
B
Observer O
Observer O
Observer B
(translation-only
Relative velocity with O)
This is an equation of adding vectors99
of different viewpoint (world) !!!
The passenger aircraft B is flying with a linear motion to theeast with
velocity vB = 800 km/h. A jet is traveling south with velocity vA = 1200
km/h. What velocity does A appear to a passenger in B ?
vA B  vA  vB
Solution
vB  800
vA B 
vA  1200
q
vA  1200 ˆj
y
vA B
800
tan q 
2
 1200 
800
1200
vB  800 iˆ

v A B  800ˆi  1200ˆj
x
100
2
Translational-only relative velocity
vA 
18 ˆ
i  5iˆ m / s
3.6
aA  3iˆ m / s2
vA B
aA B
vA B  vA  vB
aA B  aA  aB
q  2  3 
1


rad/s
60 10
q 0
You can find v and a of B101
v2
an  rq 
r
at  rq  0
2
v
vA  5iˆ m / s
q

10
aA  3iˆ m / s2
q 0
rad/s
vA
2
B
v
9
a

B
vB  rq  
R
10
9
vB  (  )  cos 45o i  sin 45o j   2iˆ  2 ˆj
10
vA/ B  vA  vB  3iˆ  2 ˆj m / s
2
B


v
aB 
 cos 45o iˆ  sin 45o ˆj  0.628iˆ  0.628 ˆj
R
aA/ B  aA  aB  3.628iˆ  0.628 ˆj m / s
y
vB
vA/B
x
Velocity Diagram
y
aB
aA
aA/B
x
Acceleration
Diagram102
Is observer B a translating-only observer
B
relative with O
vA  vB  vA/ B
?
vB ? vA  vB/ A
Yes
Yes
O
vA ? vB  vA/ B
vB ? vA  vB/ A
Yes
No
vB  vA  vrel:B / A  w 
?r
To increase his speed, the water skier A cuts across the wake of the
tow boat B, which has velocity of 60 km/h. At the instant when
q = 30°, the actual path of the skier makes an angle  = 50° with
the tow rope. For this position determine the velocity vA of the skier
and the value of q
v  rq  10 q
Relative Motion:
AB
(Cicular Motion)
20
vA B : atobserber
Consider
point A and B,
B
as r-q coordinate system
  50
vA  vB  vA B
M
?
?
Point: Most 2 unknowns can
be solved with
1 vector (2D) equation.
A
q  30
16.67
sin 40 
60
O.K.

10 m
vB / A : obserber A,
translating?
B
20
60
30
 30
D
translating?
vA B
40
60
vA
vA
vA
sin 120 

 22.5 m s
v A B  16.67
120
20
vB 
60
 16.67 m s
3.6
sin 20
 10q
sin 40
104rad s
q  0.887
2/206 A skydriver B has reached a terminal
vB  50 m / s
speed vB  50 m/s . The airplane has the constant speed
vA  50 m/s and is just beginning to follow the circular path v  50 ˆj
B
shown of curvature radius = 2000 m
aB  0
Determine
(a) the vel. and acc. of the airplane relative to skydriver.
(b) the time rate of change of the speed v r of the
vA  50iˆ
airplane and the radius of curvature r of its path, both
aA  0
observed by the nonrotating skydriver.
aA x  0  (aA )t
 a A  y  (a A )n 
v A2
A
a A  (a y ) ˆj  1.250 ˆj m / s 2
vA / B = vA - vB , aA / B  aA - aB
rB / A ,qB / A
vA/ B  50iˆ  50 ˆj
aA / B  1.250 ˆj
105
(b) the time rate of change of the speed v r of the
airplane and the radius of curvature r of its path,
both observed by the nonrotating skydriver.
vB  50 ˆj
aB  0
vA  50iˆ
aA  1.250 ˆj m / s2
vA / B , aA / B
t
aA / B
n
v r r
n  t coord
45o
vA/ B
45o
vA/ B  50iˆ  50 ˆj
aA / B  1.250 ˆj
vr  (aA/ B )t  aA/ B sin 45o
vA2 / B
r
 (aA/ B )n  aA/ B cos 45o
106
vA 
1000 ˆ
i m/s
3.6
aA  1.2iˆ m / s2
1500 ˆ
vB 
i m/s
3.6
aB  0 m / s2
r ,q : relative world
rB / A ,qB / A
r q
coord
vB / A , aB / A
107
vA 
1000 ˆ
i m/s
3.6
aA  1.2iˆ m / s2
 30o
vB 
q
r
v
a
vB / A
500 ˆ

i
3.6
q
r  q coord
r
( vB / A ) r  r
 v cos q
(vB / A )q  rq
 v sin q
(aB / A )r  r  rq 2
(aB / A )q  rq  2rq
aB / A  1.2iˆ
  a cos q
 a sin q
1500 ˆ
i m/s
3.6
aB  0 m / s2
1800  1200
 1200
sin 30o
r  v cos q  120.3
q  0.00579
r  0.637
q  0.166  103
108